Conflict-Free Colouring using Maximum Independent Set and Minimum Colouring

12/04/2018 ∙ by S. M. Dhannya, et al. ∙ 0

We present a polynomial time reduction from the conflict-free colouring problem in hypergraphs to the maximum independent set problem in simple graphs. Specifically, we show that the conflict-free colouring number of a hypergraph with m hyperedges is k if and only if the simple graph output by the reduction, denoted by G_k, has a maximum independent set of size m. We show that the simple graph resulting from this reduction applied to an interval hypergraph with three disjoint intervals is a perfect graph. Based on this we obtain a polynomial time algorithm to compute a minimum conflict-free colouring of interval hypergraphs, thus solving an open problem due to Cheilaris et al. We also present another characterization of the conflict-free colouring number in terms of the chromatic number of graphs in an associated family of simple graphs. We use this characterization to prove that for an interval hypergraph the conflict-free colouring number is the minimum partition of its intervals into sets such that each set has an exact hitting set (a hitting set in which each interval is hit exactly once).

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1 Introduction

A colouring of the vertices of a hypergraph is called a conflict-free colouring if every hyperedge has at least one vertex that has a colour different from other vertices in . Following Smorodinsky et al. [5], we abbreviate conflict-free to CF in this paper. The minimum number of colours with which a hypergraph can be CF coloured is the CF colouring number of the hypergraph and is denoted by . Computing the CF colouring number and a corresponding colouring of a given hypergraph is the CF colouring problem. Motivated by a frequency assignment problem in cellular networks, Even, Lotker, Ron and Smorodinsky [12] introduced this problem and published the first paper on CF colouring. The survey due to Smorodinsky [22] presents a general framework for CF colouring a hypergraph. This framework involves finding a proper colouring in every iteration and giving the largest colour class a new colour. They showed that if for every induced sub-hypergraph , the chromatic number of is at most , then , where . Pach and Tardos [20] have shown that if edges for some positive integer , and is the maximum degree of a vertex in , then and . Since its inception, the CF colouring problem has been studied on different types of hypergraphs. Even et al. [12] have studied a number of hypergraphs induced by geometric regions on the plane including disks, axis-parallel rectangles, regular hexagons, and general congruent centrally symmetric convex regions in the plane. Smorodinsky showed that for the hypergraph, , induced by a set of finite discs in the plane, then [21]. Similarly, if is a set of axis-parallel rectangles in the plane, then, . There have been many studies on hypergraphs induced by neighbourhoods in simple graphs. Given a simple graph , the open neighbourhood (or simply neighbourhood) of a vertex is defined as follows: . The set is known as the closed neighbourhood of . Pach and Tardos [20] have shown that the vertices of a graph with maximum degree can be coloured with colours, so that the closed neighbourhood of every vertex in is CF coloured. They also showed that if the minimum degree of vertices in is , then the open neighbourhood can be CF coloured with at most colours. Abel et al. [1] gave the following tight worst-case bound for neighbourhoods in planar graphs: three colours are sometimes necessary and always sufficient. Keller and Smorodinsky [18] studied conflict-colourings of intersection graphs of geometric objects. They showed that the intersection graph of pseudo-discs in the plane admits a CF colouring with colours, with respect to both closed and open neighbourhoods. Ashok et al. [3] studied an optimization variant of the CF colouring problem, namely Max-CFC. Given a hypergraph and integer , the problem is to find a maximum-sized subfamily of hyperedges that can be CF coloured with colours. They have given an exact algorithm running in time. The paper also studies the problem in the parametrized setting where one must find if there exists a subfamily of at least hyperedges that can be CF coloured using colours. They show that the problem is FPT and have given an algorithm with running time .

Another line of results in the literature focus on discrete interval hypergraphs (refer Section 1.1 for definition)[12]. It was shown that a hypergraph formed by all intervals on a set of points on a line can be CF coloured using colours. Chen et al.[7] presented results on an online variant of this problem, where A point has to be assigned a colour upon its arrival and the resulting colouring should be conflict-free with respect to all intervals. They gave a greedy algorithm that uses colours, a deterministic algorithm that uses colours and a randomized algorithm that uses colours. The case when interval hypergraphs have a given set of intervals as the hyperedges has been of interest [5, 17]. Katz et al.[17] gave a polynomial time algorithm for CF colouring an interval hypergraph with approximation ratio 4. Cheilaris et al.[5] improved this result in their paper on -strong CF colouring problem. The authors gave a polynomial-time approximation algorithm with approximation ratio 2 for and , when . Further, they presented a quasipolynomial time algorithm for the decision version of the -strong CF colouring problem.

Our Results: We consider the open problem from Cheilaris et al. [5] of whether the CF colouring of a given subset of intervals can be optimally solved in polynomial time. We show that it is -time solvable based on a fundamental understanding of CF colouring which is the central theme of this paper. For a hypergraph , and we associate a simple graph which we call a conflict graph (see Section 2 for the definition). The following characterization establishes a relationship between conflict graph of a hypergraph and its CF colouring number. Let be a hypergraph with hyperedges, , and denote the conflict graph of . Let be the smallest for which the size of maximum independent set in is . Then, . Further, can be constructed in polynomial time and thus the CF colouring problem on hypergraphs can be reduced in polynomial time to the maximum independent set problem in simple graphs. We believe that based on this reduction we can study CF colouring problems on different geometric hypergraphs, given that there is a vast understanding of bounds on the maximum independent sets of simple graphs like the Turán’s theorem[2]. Further we show that for interval hypergraphs which contain three disjoint hyperedges, the conflict graphs are perfect graphs. Using this we obtain a polynomial time algorithm to find the CF colouring number of interval hypergraphs, thus solving the open problem posed by Cheilaris et al.[5]. We believe that even for the hypergraph where the hyperedges are paths in a tree and the vertices are the tree vertices, the same result can be obtained using similar proofs as the ones we present in this paper. Let be an interval hypergraph such that there are at least 3 pairwise disjoint intervals in . For , the conflict graph of is a perfect graph. By combining the results in Theorems 1 and 1, in Section 2 we give an optimal algorithm to solve the CF colouring problem in interval hypergraphs. The CF colouring problem in interval hypergraphs can be solved in polynomial time. We also present results connecting CF Colouring problem and the Exact Hitting Set (EHS) problem. The EHS problem is the NP-hard dual of the Exact Cover problem [11], which is one of the earliest known NP-hard problems. For a hypergraph , we observe that a CF colouring of using at most non-zero colours partitions into hypergraphs such that each hypergraph has an exact hitting set. This is a very natural observation and we present a proof of this observation in Lemma 3.1. The interesting question is whether a hypergraph which can be partitioned into hypergraphs, each of which has an exact hitting set, can be CF coloured with at most colours. We answer this question in the affirmative in two cases: when (in Lemma 1.1) and in the case when the hypergraph is an interval hypergraph. Our results are based on the following characterization of the CF colouring number in terms of the chromatic number of a co-occurrence graph (defined in Section 3) associated with a given hypergraph. Let be a hypergraph. Let be the number of non-zero colours used in any optimal CF colouring of . Let be the minimum chromatic number over all possible co-occurrence graphs of . Then, . Based on this theorem we believe that it is possible to obtain interesting bounds on the CF colouring number by showing the existence of corresponding co-occurrence graphs of bounded degree or other properties that give a bound on the colouring number. In this spirit, we are able to show that the co-occurrence graphs of interval hypergraphs are perfect in Section 3.1. The co-occurrence graphs of interval hypergraphs are perfect. Using Theorem 1 and Theorem 1, in Section 3 we prove the following theorem, and we do not know if a similar statement holds for other geometric hypergraphs. For an interval hypergraph , there exists a partition of into parts such that for each , is an exactly hittable hypergraph if and only if there exists a CF colouring of with non-zero colours.

1.1 Preliminaries

We use to denote the degree of a vertex , and other definition and notations are standard from West [23] and Smorodinsky [22]. A hypergraph , where and is known as a discrete interval hypergraph [6]. Naturally, a hyperedge in is called an interval. A hypergraph such that the set of hyperedges is a family of intervals is known as an interval hypergraph. In an interval , and are the left and right endpoints of respectively, denoted by and , respectively. Since an interval is a finite set of consecutive integers, it follows that is well-defined.

Let be a hypergraph and be a function, which we refer to as a vertex colouring function. is defined to be a CF colouring [6] of using colours if for every hyperedge there exists a non-zero colour such that . If vertex has been assigned a colour different from the colour of all other vertices in , then we say that is CF coloured by .

A hitting set of is a set that has at least one vertex from every hyperedge. If every hyperedge is intersected exactly once by , then is called an exact hitting set. We refer to a hypergraph which has an exact hitting set as an exactly hittable hypergraph. The following lemma, used in Theorem 1, shows that a CF colouring using one colour is equivalent to an exact hitting set. A hypergraph has a CF colouring using one colour if and only if is an exactly hittable hypergraph. If is an exactly hittable hypergraph, then we get a natural CF colouring with one non-zero colour by giving the non-zero colour to the vertices in the exact hitting set, and the colour 0 to all other vertices. Similarly, in a CF colouring using one colour, each hyperedge contains exactly one vertex which is assigned the non-zero colour, and these vertices form an exact hitting set. Hence the lemma. While the recognition of exactly hittable hypergraphs is well-known to be hard, there are polynomial time recognition algorithms for the special case of exactly hittable interval hypergraphs [10, 19]. The following theorem is used in the proof of Theorem 1. [Theorem 4 in [19]] There exists a polynomial time algorithm which decides if an interval hypergraph is an exactly hittable hypergraph. Perfect Graphs: A simple graph is said to be perfect if the chromatic number, denoted by , of every induced subgraph of equals the clique number, denoted by , of . The clique number and the independence number of perfect graphs can be computed in polynomial time [15, 16], and the ellipsoid method based algorithm was first presented in the paper by Grötschel et al. [14]. This algorithm is crucial to our main result. [Chapter 6 in [14]] A maximum independent set of a perfect graph can be computed in polynomial time. A berge

graph is a simple graph which does not have an odd hole or an odd anti-hole as an induced subgraph

[4, 8, 9, 13]. A hole is an induced cycle of length at least 5. [Strong Perfect Graph Theorem (Theorem 1.2 in [9])] A graph is perfect if and only if it is Berge.

2 Conflict Graphs, Independent sets and Conflict-free colourings

Given a hypergraph , and integer , we now formally define the conflict graph . is a simple graph that encodes the constraints to be respected by a CF colouring of with at most non-zero colours. The vertex set of is . The elements of are referred to as nodes and the word vertex refers to a vertex of a hypergraph. In a node , we refer to as the hyperedge coordinate, as the vertex coordinate and as the colour coordinate. Conceptually, a node in represents the logical proposition that hyperedge is CF coloured by vertex which has been assigned the colour . We reiterate that in a node , the vertex is an element of the hyperedge . The edge set of is a subset of 2-tuples of . is defined such that each edge encodes a constraint to be respected by any CF colouring of . The edge set of is , where and are defined as follows:

  1. . Note that the set of nodes in whose vertex coordinate is forms a complete -partite graph.

  2. . For each hyperedge in , the nodes in with as the hyperedge coordinate are all adjacent.

  3. . For each colour , an edge is created for those such that either and both are in or both in .

Figure 1 gives an example of an interval hypergraph and its conflict graph when . In Figure 1(ii), the solid edges belong to , the dashed edges belong to and the dotted edges belong to . In order to avoid cluttering, not all edges in have been shown. There is a natural correspondence between the set of independent sets of and the set of vertex colourings of with colours.

clique of

clique of

clique of

Figure 1: (i) An interval hypergraph (ii) Conflict graph of for

Independent sets in and vertex colourings of : Given an independent set of , consider the following vertex colouring function defined as follows. For each ,

Next, given a vertex colouring function of , we define a subset of nodes in as follows. Consider the set of hyperedges in , denoted by , that are CF coloured by . For each hyperedge , let be one (any one, if there are more than one) node such that is CF coloured by and . Let denote the set containing all these nodes. We call the conflict-free set obtained from . In the following two lemmas, we formally prove the connection between a CF colouring of and a maximum independent set in by proving properties of vertex colouring functions and conflict-free sets.

Notation: In the rest of this section refers to a conflict graph and the corresponding and will be clear from the context. Similarly, the conflict graph associated with a vertex colouring function obtained from an independent set will be clear from the context. Finally, the vertex colouring function associated with a conflict-free set in the hypergraph will be clear from the context. For an integer and a hypergraph , let be the conflict graph of and let . Then the vertex colouring function obtained from any maximum independent set in is a CF colouring of with at most non-zero colours. Let be a maximum independent set of . By definition of the edges in , it follows that for all the nodes in , the hyperedge coordinates are all distinct. For a hyperedge , the nodes in with as the hyperedge coordinate form a clique in . Therefore, all the nodes in have distinct hyperedge coordinates. Consequently, for , let be the node in . From the definition of the edges in , we observe that the vertex colouring function defined based on is indeed a function on such that the range of is a subset of . This follows from the fact that if and are two nodes in such that , then . Indeed, if and were distinct colours, then by the definition of , there should have been an edge which does not exist as the two nodes are in the independent set . We next show that is a CF colouring of : Let . Let be another vertex in . We now show that is different from . If , then clearly and are different, since is a non-zero value. In the case when is non-zero, then by the definition of there is a node in such that the vertex coordinate is the vertex . Since and are distinct elements of the hyperedge , and because and are nodes in the independent set and thus non-adjacent in , it follows that . The reason is that had they been equal, by the definition of , would have been an edge in , which would contradict the fact that and are nodes in the independent set . Therefore, and are different. Consequently, it follows that for any in different from , and are different. Hence is a CF colouring of . Since the range of has at most non-zero colours, it follows that is a CF colouring of using at most non-zero colours. Hence the lemma. We next show that a conflict-free set in obtained from a CF colouring using at most non-zero colours is a maximum independent set in . Let be a CF colouring of that uses non-zero colours. Then, a conflict-free set obtained from is a maximum independent set of the conflict graph . Since is a CF colouring of , it follows by the definition of that for each , there is exactly one node in for which the hyperedge coordinate is . Therefore, . Let . We now show that is an independent set. Let and be two nodes in . By the construction of , it follows that . We consider two cases: in the case when , then since they are colours given by to . From the definitions of and , it follows that and are non-adjacent. In the case when , we have two sub-cases here: in the case when and do not both belong to and do not both belong to , then by the definition of , it follows that and are non-adjacent. In the sub-case when both and belong to either or , then since is a conflict-colouring of in which and are CF coloured by and , respectively, it follows that . From the definition of , in this case it follows that and are non-adjacent. Therefore, is an independent set in . Since the nodes of are partitioned into at most cliques (one clique for each hyperedge ), it follows that the maximum independent set in has at most nodes. Therefore, is a maximum independent set of size in . Hence the lemma.

[Proof of Theorem 1:] The proof of this theorem follows from Lemmas 1 and 2. Let be a maximum independent set of . Given . Let be the vertex colouring function obtained from . It follows from Lemma 1 that is a CF colouring of with at most colours. Hence .
To show the other direction, consider any optimal CF colouring of . Let be the number of colours used by . That is, . Let be the conflict-free set obtained from . It follows from Lemma 2 that is a maximum independent set of graph and it is of size . Therefore, . Therefore, it follows that .
Also for each hypergraph and , it follows from the description of that it can be constructed in polynomial time. Further, since it follows that we have a polynomial time reduction from the CF colouring problem in hypergraphs to the maximum independent set problem in simple graphs. Hence the theorem.

2.1 Conflict Graph of an Interval Hypergraph is Perfect

In this section, we show that the conflict graph of an interval hypergraph is a perfect graph when there are at least three disjoint intervals in . We use Theorem 1.1 to show this perfectness result. For an interval hypergraph and , let denote the number of vertices in . Note that . [Proof of Theorem 1:] By the characterization of perfect graphs in Theorem 1.1, we know that for each , induced odd cycle and its complement denoted by are forbidden induced subgraphs. We now show that for each integer and an interval hypergraph which has at least three disjoint intervals, the graph is perfect. Our proof is by starting with the hypothesis that the claim is false and deriving a contradiction. For a , let be an interval hypergraph for which is imperfect, and among all such interval hypergraphs, minimizes . Since is imperfect, let us consider a minimal imperfect subgraph of , denoted by . We first observe that for every interval , there exists colours such that and belong to . The proof of this claim is by contradiction to the fact that is an interval hypergraph that minimizes . Let be an interval in such that for all colours , the node is not a node in . Consider the hypergraph where . Let denote the conflict graph of . The only difference between and is that the set is not a subset of . In other words, . However, none of the nodes in are elements of , and therefore is an induced subgraph of . Therefore by Theorem 1.1 it follows that is imperfect. Further, . This contradicts the hypothesis that is the interval hypergraph with minimum for which is imperfect. Therefore, it follows that for each interval , there exists a colour such that is a node in , the minimal imperfect graph in . The same argument shows that for each interval , there exists a colour such that is a node in , the minimal imperfect graph in . Hence it follows that , there exists colours such that and belong to . We now consider two exhaustive cases to obtain a contradiction to the known structure of the minimal imperfect subgraph which we know is either a or a for some .
Case 1- When is an induced odd cycle : In the preceding argument, we showed that for each interval , there exist such that and belong to . This accounts for an even number of distinct nodes in the cycle . It follows that each interval contributes even number of nodes to . Since is an induced odd cycle, it follows that in there is at least one more node for which is different from and . From the definition of , we know that the 3 nodes form a . This is a contradiction to the fact that an induced cycle of length at least 5 does not have a as an induced subgraph. Therefore, the minimal imperfect subgraph is not an induced odd cycle.
Case 2- When is the complement of an odd cycle, say : Consider 3 pairwise disjoint intervals in . We have already shown that for each interval , there exist such that and belong to . It follows that there exists colours such that , , , , , belong to . Since are pairwise disjoint, it follows from the construction of the conflict graph that for all , there is no edge from to and there is no edge from to . It follows that form an independent set. It follows that there is an independent set of size at least 3 in . This is a contradiction to the fact that in the complement of any induced cycle of length at least 4, there is no independent set of size greater than 2. Therefore, cannot be the complement of an induced odd cycle of length at least 5.
As a consequence of the above analysis, we observe that starting with the hypothesis that is imperfect, we get a contradiction to the known structure of a minimal imperfect subgraph . Therefore, our hypothesis that is imperfect is wrong. Hence, it follows that for and an interval hypergraph with at least 3 disjoint intervals, is perfect. Hence the theorem. Having shown that the conflict graph of an interval hypergraph with three disjoint intervals is perfect, we next bound the CF colouring number of an interval hypergraph which does not contain 3 pairwise disjoint intervals. Let be an interval hypergraph which does not contain three disjoint intervals in . Then . Further, such an interval hypergraph can be recognized in polynomial time. Let us consider the intersection graph of the set of intervals which we know is an interval graph. It is well-known, see for example the book by Golumbic [13], that the interval graph is perfect. Since there do not exist 3 disjoint intervals, it follows that the interval graph has a maximum independent set of size at most 2. From the definition of perfect graphs (see Section 1.1), we know that the size of the maximum independent set is equal to the size of the minimum clique cover, and it can be found in polynomial time. Therefore, the interval graph of has a clique cover of size at most 2. The clique cover gives two corresponding points in that intersect each interval in . We now consider the following vertex colouring function defined on : colour one of the points with colour 1 and the other point with colour 2, and all the other points are coloured 0. Since the colours 1 and 2 are given to exactly two vertices in and all the other vertices are given the colour 0, this vertex colouring function is a CF colouring of . Thus, and the recognition of such interval graphs can also be done in polynomial time. We now prove Theorem 1 which is an application of our reduction from the CF colouring problem on hypergraphs to the maximum independent set problem in simple graphs. [Proof of Theorem 1:] If is an exactly hittable interval hypergraph then, by Lemma 1.1, . From Theorem 1.1, an exactly hittable interval hypergraph can be recognized in polynomial time. If is not exactly hittable and if does not have 3 disjoint intervals, then by Lemma 2.1, and such an can be recognized in polynomial time. If the clique cover of is , then there are at least three pairwise disjoint intervals in . In this case, we reduce the CF colouring problem in to the maximum independent set problem in the conflict graph . We iterate for values of starting from and find the smallest , say , for which the independence number of conflict graph is . By Theorem 1, . Observe that by Theorem 1, for each , the conflict graph is perfect. By Theorem 1.1, the independence number of perfect graphs can be obtained in polynomial time. Additionally, it is known from Theorem 1 that the conflict graph can be constructed in polynomial time. From the above results it follows that an optimal CF colouring of can be obtained in polynomial time.

3 Conflict-free Colouring and the Chromatic Number of Co-occurrence Graphs

Given a hypergraph , we show that the CF colouring number of is the minimum chromatic number over a set of simple graphs associated with . In this section we define the simple graph which is called a co-occurrence graph.

For a CF colouring function defined on , let be a function such that is CF coloured by . We refer to as a representative function obtained from the colouring . We now show that given a function such that for each edge , , we can obtain a CF colouring of for which is the representative function. Let denote the image of under the function . The vertex set of the co-occurrence graph is , and for , is an edge in if and only if for some , and and is either or . Note that for a conflict graph , the subscript is a positive integer, and in the case of the co-occurence graph , is a function on the vertex set . An example is given in Figure 2.

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Fig (a)

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Fig (b)
Figure 2: (a) Interval Hypergraph (b) Co-occurrence graph of with , and for each marked as the label for each interval

Define where is the chromatic number of the co-occurrence graph and the minimum is taken over all representative functions . [Proof of Theorem 1:] Let be a representative function such that . We extend a proper colouring of to a vertex colouring function of by assigning the colour to those vertices in . is a CF colouring of since for each , the colour assigned to the vertex by is different from the colour assigned to every other vertex in . The reason for this is as follows: let be a vertex different from . If , then definitely its colour is different from . On the other hand, if is non-zero, then it implies that there is an such that . Consequently, , and since , is an edge in by the definition of . Further, since is obtained from a proper colouring of it follows that is different form . Thus . We prove that as follows: since a minimum CF colouring of gives a representative function as defined above, it follows that . Therefore, it follows that . Hence the theorem.

The co-occurrence graphs of intervals are perfect

As in the case of conflict graphs, we show that the co-occurrence graphs of interval hypergraphs are perfect. [Proof of Theorem 1:] We use Theorem 1.1 to prove this perfectness result. Given an interval hypergraph , let be a representative function and let be the resulting co-occurrence graph. We first show that does not have an induced cycle of length at least 5. Note that, we prove a stronger statement than required by Theorem 1.1 which requires that there are no induced odd cycles of length at least 5. Our proof is by contradiction. Assume that is an induced -cycle for . Let the sequence of nodes in be . Let be the rightmost point of on the line. In what follows, the arithmetic among the indices of is  . Without loss of generality, let us assume that , which are the two neighbours of in . Therefore, . Since edge is in , it follows that there exists an interval for which . We claim that is : if is , then is an edge in by definition. Therefore, is a chord in , a contradiction to the fact that is an induced cycle. Therefore, . Further, we claim that the point : if , then belongs to the interval and by the definition of the edges in , is an edge in . Therefore, is a chord in . This contradicts the fact that is an induced cycle. Therefore, . At this point in the proof we have concluded that and . Since is an edge in , it follows that there exists an interval such that both and belong to and . Since is an induced cycle of length at least 5, is an edge in by definition. Therefore, chord in either case, that is when or . This contradicts the assumption that is an induced cycle of length at least 5. Thus, cannot have an induced cycle of size at least 5.

Next, we show that does not contain the complement of an induced cycle of length at least 5. Assume that is an induced , in . Let be the nodes of . Also, let be the left to right ordering of points on the line corresponding to vertices of . Since for all in , it follows that no interval , such that , contains more than vertices from . Otherwise, if there exists an interval such that contains more than vertices from , then in which is a contradiction. Therefore, there does not exist any interval that contains both and . Similarly, there does not exist any interval that contains both and and any interval that contains both and . Since , it follows that must be adjacent to all vertices in . Similarly, must be adjacent to all vertices in . Next, we consider the degrees of vertices and in . Since they are in , is adjacent to and is adjacent to . Now, must be adjacent to more vertices. We show that is not adjacent to . Suppose not, that is, if is adjacent to , then there exists an interval that contains both and and . Then is adjacent to all points in the set . Thus, by considering the one additional edge incident on depending on whether or , it follows that , a contradiction to the fact that the degree of each vertex inside is . Therefore, it follows that does not exist in . It follows that in , which we know is an induced cycle of length at least 5, there is an induced cycle of length . This contradicts the structure of an induced cycle of length at least 5. Hence, we conclude that does not have an induced cycle of length 5 or more or its complement. Therefore is a perfect graph. Hence the theorem.

3.1 Partition into Exactly Hittable Sets and Conflict-free Colouring

In this section, using the following two lemmas we prove Theorem 1. We first show that a CF colouring with at most non-zero colours partitions a hypergraph into at most exactly hittable hypergraphs. If there exists a CF colouring of a hypergraph with non-zero colours, then there exists a partition of into parts such that each is an exactly hittable hypergraph. Given a CF colouring with at most non-zero colours, let be a representative function such that for each , is CF coloured by the vertex . The hyperedges are partitioned into sets based on and the vertex colouring as follows: the set consists of all those hyperedges such that the colour of is . We show that for each , is an exactly hittable hypergraph and the exact hitting set is . is a hitting set of because for each , is in . Further, each is hitting exactly once by because all the vertices of have the same colour in the CF colouring . Thus, is an exact hitting set of . Therefore, each is an exactly hittable hypergraph. This proves the lemma. Hence the lemma. We next set up the machinery to conclude that if we are given a partition of an interval hypergraph into exactly hittable interval hypergraphs, then we get a CF colouring with at most non-zero colours. Let be a partition of intervals in , such that each is an exactly hittable interval hypergraph. We show that there is a conflict free colouring of with colours. Let be the exact hitting sets of the parts respectively. Let denote the set . For each interval , let be the only vertex in . Let be the co-occurrence graph of . In the arguments below, the graph and the representative function are as defined here. We now prove Lemmas 3.1 and 3.1 and use them in the proof of Theorem 1. Let = be a clique of size in the co-occurrence graph . Then, there are distinct parts in the set containing intervals respectively, satisfying the following property: for each in , is the representative of interval and for each edge in either is in or is in . The proof is by induction on the size of the clique. The claim is true for base case when ; then is the representative of some interval in some part . Assume that the claim is true for any clique of size . Now, we show that the claim is true for clique of size . Let be the left to right ordering of the points (on the line) corresponding to vertices in the clique . Since is an edge in , there must exist an interval such that either or is the representative of and occurs along with inside . Without loss of generality, assume that is the representative of interval . Observe that